Frame bundle

Matterwave

Hmmm, can you elaborate a bit on that?

I can see that if H is arbitrary, then my definition of what is a horizontal lift is aribtrary, and that would mean arbitrariness in parallel transport. But I can't see how H would be arbitrary, doesn't it have to encompass every "direction" that V doesn't in order to be a compliment to it?

I mean dim(V)=dim(G) and dim(H)=dim(M) and dim(P)=dim(G)+dim(M) right, so I'm not seeing how it can be arbitrary.

Perhaps a simple example would help? Or perhaps one could try to explain, for example, if my manifold is now endowed with a metric, how that limits my choice of H to be compatible somehow.

quasar987

There are always an infinite number of ways to choose a complementary subspace. The simplest example is in R². Set V:=R x {0}. There is an infinite number of ways to chose a subspace complementary to V since in fact, any subspace other than V itelf does the job, so the complementary subspaces H are in 1-1 correspondence with the angle θ in (0,pi) that they make with the horizontal.

This illustrates the situation in the linear setting. In the vector bundle setting there is even more freedom in the choice of a complementary subbundle H.

Matterwave

Ah, that makes things a lot clearer. Thanks.

EDIT: Just out of curiosity, are you a mathematician? Or a physicist who learned a lot of math? hehe...

lavinia

Quote by Matterwave

As a follow up, frame bundles, and principle bundles in general, are not vector bundles right? Since GL(n,R) doesn't seem to form a vector space. I'm a little confused because this book I'm reading is discussing the "distribution on P of vertical vectors...", where P is a principle bundle, but it seems to me that a principle bundle is not a vector bundle so where are the vectors?

The vertical vectors are tangent to the manifold of frames in each fiber. They are the kernel of the differential of the bundle projection map.

Horizontal vectors are tangent to the bundle of frames. They vary with the connection and are not in the kernel of the projection.

quasar987

Quote by Matterwave

Ah, that makes things a lot clearer. Thanks.

EDIT: Just out of curiosity, are you a mathematician? Or a physicist who learned a lot of math? hehe...

I am a grad student in math, but my undergrad degree is in physics. What about yourself? I think you said you are a physics student, but what are you doing specifically? What is your motivation for learning about all this bundle nonsense? :)

Matterwave

The book I'm using defines parallel transport along a curve c(t) in M of a vector in an associated vector bundle (taken to be TM for simplicity) as follows. Given the curve c(t) in M, horizontally lift this curve into the frame bundle giving you a curve c'(t) in the frame bundle such that c'(0)=b, where projb=c(0). This curve c'(t) gives you the parallel translate of a the point b along the curve c(t), and gives you, at each point, the parallel translated basis vectors for our associated tangent bundle {f(t)}. The parallel transport of a vector in the tangent bundle along this curve c(t) is then basically to keep the components of our vector constant as we change our basis vectors so that they are parallel transported. I.e. v(t)=sum(v(0)f(t)) where v(t) is the parallel transported vector, and v(0) are the components of this vector as t=0.

This construction makes sense to me. In essence, I am just parallel transporting my basis vectors using this connexion H on the frame bundle, and that gives me a notion of parallel transporting my vectors in TM by keeping their components constant since my basis vectors are parallel. This means that my connexion H on my frame bundle induces a connexion H' (equiv to a law of parallel transport) on my tangent bundle.

My question is, is there some (not ridiculously difficult) way to link this "new" construction of parallel transport, etc., with the "regular" construction I'm used to in the sense of a covariant derivative? I.e. is there some way to pinpoint which H I should choose, if I have to choose, i.e. on a Riemannian manifold the unique law of parallel transport that preserves inner products and has no torsion? This should pinpoint for me a unique distribution H in my frame bundle right? But how do I know which H?

EDIT: I have an undergrad degree in Astrophysics and Economics, and am a graduate student in physics. I'm learning this mostly for my own enjoyment because I have a lot of interest in differential geometry, but also because I want a clear mathematical understanding of the concepts I use in general relativity and quantum field theory for connections, etc. =]

quasar987

You can look at Chris Wendl's remarkably enlightening lecture notes on bundles (see his website). Particularly, Theorem 3.38 exhibits the link between connexions on the frame bundle and the corresponding connexions in the associated vector bundle.

here:http://www.homepages.ucl.ac.uk/~ucah...nnections.html

Matterwave

Ok, I'll take a look thanks =]

lavinia

An Example: The torus is a circle bundle over the circle and has a natural action of SO(2) on each fiber - just rotate. Any vector field that is invariant under fiber rotation and which is never tangent to any fiber circle generates a horizontal space. The connection one form is easy to define from the decomposition of a vector into horizontal and verticle components.

to see that there are infinitely many connections, just take any smooth vector field on a circle that is transverse to the fiber and rotate is around to get a vector field on the whole torus.

lavinia

An example: (Please correct this if it is wrong.)

On a surface (two dimensional Riemannian manifold)

Take a curve that is parameterized by arc length and a unit vector field ,v,along the curve that is perpendiclular to the curve. The covariant derivative of v will be tangent to the curve and so is some multiple of c'(s).

To compute this multiple view v as a map of the curve into the tangent circle bundle (same as the bundle of orthonormal frames). dv(c'(s)) is the derivative of v with respect to c'(s).

Decompose dv(c'(s)) into horizontal and vertical components. The multiple we are looking for is the coefficient of the vertical component of this decomposition.

c'(s )is also a map of the curve into the tangent circle bundle and the argument is the same.
The covariant derivative this time will be in the v direction normal to c'(s).

Extend now by the Leibniz rule and linearity to arbitrary v.

Bacle2

Quote by quasar987

I will try to clear things up a little for you, if that is possible.

I wouldn't put it that way, because it sounds like you're saying that fibers are made up of basis for the vector space of the vector fields on M, which is false of course. More precise would be that the frame bundle over a manifold M is a principle GL(n,R)-bundle who's fibers are the sets of ordered bases for the tangent space of M at p. (where n=dim(M))

Yes, it is literally an n-tuple of vectors spanning T_mM.

You have to be careful here because a fiber bundle (such as a principal bundle) is itself a manifold. So you have to distinguish the dimension of the bundle as a manifold and the dimension of its fibers. Here, 16=4² is the dimension of the fiber. The dimension of the frame bundle as a manifold is 16+4=20. What do you mean "why so many dimensions"? I would think that you know the answer to that since you computed correctly that GL(4,R) has dimension 16!

Ah, perhaps you do not understand what the connection is between the fact on the one hand that the fiber over m are the frames of TmM and the claim that this space is just GL(4,R). This is simply because in a vector space V of dimension n, once you fix a basis (v_i), you can then express any vector w in V as a set of n real numbers; namely the so-called coordinates of w wrt to (v_i):
[tex]w=\sum_i a_iv_i[/tex]
In particular, if (w_j) is another basis of V, then this corresponds to n² numbers (n numbers for each w_j), which you can arrange in a matrix by declaring that the first row is to be made up of the n coordinates of w₁ and so on. Then the fact that these (wj) are linearly independant is equivalent to saying that the matrix thus constructed has nonvanishing determinant. That is, it is a matrix in GL(n,R). So you see that there is a (non canonical) bijective correspondance between the frames F(V) of V and GL(n,R). Use that to transfer the smooth structure of GL to F(V). You can verify that this will be independant of the choice of bijection and so puts a well-defined canonical smooth structure on F(V) such that given any choice of basis in V resulting as above in a bijection F(V)<-->GL, this bijection is a diffeomorphism. So that is how each fiber of the frame bundle is (non canonically) diffeomorphic to GL(n,R).

Quasar:
When you refer to the dimension of Gl(4,R) ; algebraic dimension, if I understood well, is equal to 4^2=16. What kind of algebraic structure are you assigning to Gl(4,R)?

morphism

GL(n,R) is an open subset of ##M_n(\mathbb R) \cong \mathbb R^{n^2}##, so it's a manifold of dimension equal to dim M_n(R) = n^2.

Bacle2

I understand that, but I thought I read a reference to algebraic dimension, tho I may have misunderstood--or misunderestimated what I read.

quasar987

No reference to algebraic dimension. :)

Bacle2

Sorry, Quasar,
I misunderestimated your question :). Nice post, btw.

Bacle2

Sorry again for my slowness, Quasar. I understand that Gl(n,R)=Det^-1(ℝ\{0}), which is open, yada, yada; I realized where my confusion lay.

Anyway, another dumb question; I am also trying to teach myself some bundles: (my apologies, my quoting button is disabled for some reason): say you are considering the case of a vector bundle ( thinking mostly of tangent bundle), in a situation where you have a metric defined. It would then make sense to define the complement bundle to be the ortho-complement (with respect to this metric), right? Is there anything special to this choice?

quasar987

I'm not sure I follow you. You are thinking of the case where (M,g) is a riemannian manifold. Then you have the projection map pr:TM-->M and its differential pr_*:T(TM)-->TM and you want a complementary subbundle to V:=ker(pr_*). You cannot just pullback g to T(TM) via pr, and use that to define H:=V[itex]^{\perp}[/itex] because pr^*g is very degenerate: pr_* vanishes on V and so V[itex]^{\perp}[/itex]=T(TM).

Is this what you were thinking?

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Matterwave	Ah, that makes things a lot clearer. Thanks. EDIT: Just out of curiosity, are you a mathematician? Or a physicist who learned a lot of math? hehe...

quasar987 Recognitions: Gold Member Homework Help Science Advisor	You can look at Chris Wendl's remarkably enlightening lecture notes on bundles (see his website). Particularly, Theorem 3.38 exhibits the link between connexions on the frame bundle and the corresponding connexions in the associated vector bundle. here:http://www.homepages.ucl.ac.uk/~ucah...nnections.html

morphism Recognitions: Homework Help Science Advisor	GL(n,R) is an open subset of ##M_n(\mathbb R) \cong \mathbb R^{n^2}##, so it's a manifold of dimension equal to dim M_n(R) = n^2.

Bacle2 Recognitions: Science Advisor	I understand that, but I thought I read a reference to algebraic dimension, tho I may have misunderstood--or misunderestimated what I read.